JEE Chemistry Redox Reactions Notes Set 03

Oxidation and Reduction

Classical view: oxidation is addition of oxygen or removal of hydrogen; reduction is removal of oxygen or addition of hydrogen. Electronic view: oxidation is loss of electrons and reduction is gain of electrons; the two always occur together (redox reaction).
Oxidising agent: accepts electrons and is itself reduced (F\(_2\), O\(_3\), KMnO\(_4\), K\(_2\)Cr\(_2\)O\(_7\), HNO\(_3\), conc. H\(_2\)SO\(_4\), Cl\(_2\)). Reducing agent: donates electrons and is itself oxidised (metals, H\(_2\), C, CO, H\(_2\)S, SO\(_2\), SnCl\(_2\), I\(^-\), FeSO\(_4\)).
A substance whose element is in its highest oxidation state can only be an oxidising agent; in its lowest state only a reducing agent; in an intermediate state it can act as both (H\(_2\)O\(_2\), SO\(_2\), HNO\(_2\), NaNO\(_2\)).

Oxidation Number

Rules: the oxidation number of an element in any form is zero; of a monatomic ion equals its charge; the sum over a neutral compound is zero and over an ion equals its charge. Alkali metals +1, alkaline earth metals +2, fluorine -1. Hydrogen is +1 (but -1 in metal hydrides like LiH). Oxygen is -2 (but -1 in peroxides such as H\(_2\)O\(_2\) and Na\(_2\)O\(_2\), -1/2 in superoxides, +2 in OF\(_2\)).
Examples: Mn in KMnO\(_4\) +7; Cr in K\(_2\)Cr\(_2\)O\(_7\) +6; S in H\(_2\)SO\(_4\) +6, in H\(_2\)S\(_2\)O\(_8\) +6, in Na\(_2\)S\(_2\)O\(_3\) +2; N in HNO\(_3\) +5, in NH\(_4\)\(^+\) -3; Cl in HClO\(_4\) +7. The maximum oxidation state of an element equals its group number (for main groups) and the minimum is group number minus 8.

Types of Redox Reactions

Combination (C + O\(_2\) \(\rightarrow\) CO\(_2\)), decomposition (2H\(_2\)O\(_2\) \(\rightarrow\) 2H\(_2\)O + O\(_2\)), displacement (Zn + CuSO\(_4\) \(\rightarrow\) ZnSO\(_4\) + Cu).
Disproportionation: the same element is oxidised and reduced, e.g. Cl\(_2\) + 2NaOH \(\rightarrow\) NaCl + NaClO + H\(_2\)O and 2H\(_2\)O\(_2\) \(\rightarrow\) 2H\(_2\)O + O\(_2\). Comproportionation is the reverse.

Balancing Redox Equations and Equivalent Weight

Ion-electron method: write the two half reactions, balance atoms other than H and O, balance O with H\(_2\)O, balance H with H\(^+\) (acid) or OH\(^-\) (base), balance charge with electrons, multiply to equalise electrons and add. Oxidation number method: equalise the total increase and decrease in oxidation number.
Equivalent weight of an oxidising or reducing agent = molar mass / number of electrons gained or lost per formula unit. KMnO\(_4\): M/5 in acid, M/3 in neutral, M/1 in strong alkali. K\(_2\)Cr\(_2\)O\(_7\): M/6 in acid. Oxalic acid or oxalate: M/2. FeSO\(_4\): M/1.

Worked Examples

Example 1. Find the oxidation number of Cr in \( [Cr(NH_3)_4Cl_2]^+ \).
Answer:
NH\(_3\) is neutral and Cl is -1: \( x + 0 + 2(-1) = +1 \), so \( x = +3 \).

Example 2. How many moles of KMnO\(_4\) are needed to oxidise one mole of Fe\(^{2+}\) in acid solution?
Answer:
MnO\(_4^-\) gains 5 electrons and Fe\(^{2+}\) loses 1.
\( MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \), so 0.2 mol KMnO\(_4\) is required.